JEE Mains · Physics · STD 12 - 3. current electricity
Two cells of emf \(2\, E\) and \(E\) with internal resistance \(r _{1}\) and \(r _{2}\) respectively are connected in series to an external resistor \(R\) (see \(figure\)). The value of \(R ,\) at which the potential difference across the terminals of the first cell becomes zero is
- A \(r _{1}+ r _{2}\)
- B \(\frac{ r _{1}}{2}- r _{2}\)
- C \(\frac{ r _{1}}{2}+ r _{2}\)
- D \(r _{1}- r _{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{ r _{1}}{2}- r _{2}\)
Step-by-step Solution
Detailed explanation
\(i =\frac{3 E }{ R + r _{1}+ r _{2}}\) \(TPD =2 E - ir _{1}=0\) \(2 E = ir _{1}\) \(2 E =\frac{3 E \times r _{1}}{ R + r _{1}+ r _{2}}\) \(2 R +2 r _{1}+2 r _{2}=3 r _{1}\) \(R =\frac{ r _{1}}{2}- r _{2}\)
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